Question

- If alpha, Beta and gamma are the zeroes of the polynomial f (x) = px^3+ 9x^2 + rx + s, then find the value of
Alpha^2 + beta^2 + gamma^2

Best answer will get brainliest
Spammer will get reported
Thanks!

Answer 1

Step-by-step explanation:

$\alpha \: \beta \: \gamma \: are \: zero \: of \: f(x) = px {}^{3} + 9 {x}^{2} + rx + s$

where a= p , b = q , c = r, d = s.

$\alpha + \beta + \gamma = \frac{ - b}{a}$

$\alpha + \beta + \gamma = \frac{ - q}{p}$

________1

αβ+βγ+αγ= c/q

⇒αβ+βγ+αγ = r/p_________ (2)

⇒αβγ= - d/q

αβγ= -s/p________ (3)

$\alpha {}^{2} + \beta + {}^{2} \gamma {}^{2} = ( \alpha + \beta + \gamma ) {}^{2} - 2 \alpha \beta - 2 \beta \gamma - 2 \gamma \alpha$

$(\alpha + \beta + \gamma ) {}^{2} - 2( \alpha \beta + \beta \gamma - \gamma \alpha )$

(q/p)^2 - 2 r/p. [ from 1 & 2 ]

q^2/p^3 - 2r/p

q^2 -2rp/p^2

.

Answer 2

Answer: q^2/p^2 - 2r/p or (q^2 - 2pr)/p^2.

Explanation:

Given, p(x) = px^3 + qx^2 + rx + s

We know,

α + β + γ = - b/a

αβ + βγ + αγ = c/a

Also,

(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac)

=> (- b/a)^2 = (α^2 + β^2 + γ^2) + 2(c/a)

=> (α^2 + β^2 + γ^2) = (- q/p)^2 - 2(r/p) = q^2/p^2 - 2r/p.